Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $17.8$ years; the standard deviation is $3.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living longer than $21.4$ years.
$17.8$ $14.2$ $21.4$ $10.6$ $25$ $7$ $28.6$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $17.8$ years. We know the standard deviation is $3.6$ years, so one standard deviation below the mean is $14.2$ years and one standard deviation above the mean is $21.4$ years. Two standard deviations below the mean is $10.6$ years and two standard deviations above the mean is $25$ years. Three standard deviations below the mean is $7$ years and three standard deviations above the mean is $28.6$ years. We are interested in the probability of a tiger living longer than $21.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $14.2$ years and the other half $({16\%})$ will live longer than $21.4$ years. The probability of a particular tiger living longer than $21.4$ years is ${16\%}$.